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Solution

Given,

$S\to AbB$

$A\to aAb\mid \lambda$

So, $A$ generates equal number of $a$ and $b$.

$A\Rightarrow a^n b^n$

Also,

$B\to bB\mid \lambda$

So, $B$ generates any number of $b$.

$B\Rightarrow b^k$

Now,

$S\Rightarrow AbB$

$S\Rightarrow a^n b^n b b^k$

$S\Rightarrow a^n b^{n+k+1}$

So, number of $b$ is always greater than number of $a$.

Therefore,

$L=\{a^n b^m:n\ge 0,m>n\}$